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QUESTION 3Suppose X, Y, Z are random variables such that E(XY) =20,E(x) =4,E(Z) = 10. Determine E(YZ) if X, Y are uncorrelated and Y,Z are independentNone of the al...

Question

QUESTION 3Suppose X, Y, Z are random variables such that E(XY) =20,E(x) =4,E(Z) = 10. Determine E(YZ) if X, Y are uncorrelated and Y,Z are independentNone of the alternatives is correct:50QUESTION 4There is 48% chance of rain today and 51% chance of rain tomorow. Assuming that the event that it rins today is independent of the cvent that it rains tOmorTow; what is the probability that thcre will be no rain today tomorow? Plcasc round thc answcr two dccimal placcs035 0.25 0.26 d.0.14 0.23

QUESTION 3 Suppose X, Y, Z are random variables such that E(XY) =20,E(x) =4,E(Z) = 10. Determine E(YZ) if X, Y are uncorrelated and Y,Z are independent None of the alternatives is correct: 50 QUESTION 4 There is 48% chance of rain today and 51% chance of rain tomorow. Assuming that the event that it rins today is independent of the cvent that it rains tOmorTow; what is the probability that thcre will be no rain today tomorow? Plcasc round thc answcr two dccimal placcs 035 0.25 0.26 d.0.14 0.23



Answers

Checking independence For each of the following situations, would you expect the random variables
$X$ and $Y$ to be independent? Explain your answers.

(a) $X$ is the rainfall (in inches) on November 6 of this year, and $Y$ is the rainfall at the same location on November 6 of next year.
(b) $X$ is the amount of rainfall today, and $Y$ is the rainfall at the same location tomorrow.
(c) $X$ is today's rainfall at the airport in Orlando, Florida, and Y is today's rainfall at Disney World just
outside Orlando.

Problem Number 50. We can't say that. And eggs and boy are not are not, um, independent The fools or a major. The reason people and we see a location. Ah, Ricin. Ah, be It was a This was being ah x and Roy. Uh huh. And me. Ah, independent don't. So because ah, both measure there in full at the same Ah, and the same location. So it's the same ones. Uh, question number c ah, also ex. And why I all right. We need been bent every cools on Marine It's on a jury and locations that are here It's other verse, but measurements are rightly, uh, to be related.

Probability that trains days for five want on the old sinful. So the odds of all the rain uh, offering is important, too, for over five is through one or five, which is for over five. I love my silence over warrant, which is for

This exercise, we refer back to exercise 16 which was the mark of chain, which had to do with how Maney umbrellas Sara head at home and in exercise 16. We had come up with a single step transition matrix here and now we're asked in part a that if Sarah has two umbrellas at home on Sunday night, what is the probability of having exactly two umbrellas at home on the next Friday night? So that is exactly five days later, five working days later and say and is equal to five. And so we're looking for the five step transition probability. I'm going from two to you, too. And, as usual, the way to find the transition probabilities for five steps is to take the single step transition matrix and raise it to the explode and five so he could do that in software. And so then this is equal to the entry to to in that matrix. Now we do have to be careful, because when we came up with this transition matrix and exercise 16 I had indexed the rows and columns as follows. Just so, it was easier to keep track of how many umbrellas we're at home. So here, when I say entry to to actually mean this entry right here, which is entry, actually entry 33 in in conventional matrix terminology and in the five step transition matrix, that number happens to be 0.274 So that's that's sort of the first question of part. And the second question is what is the probability by a Friday evening that there will be at least two umbrellas at the house? So this time we're looking for the five step probability going from 2 to 2, unless the five step probability of going from 2 to 3, as well as from 2 to 4. And so this is equal to zero point 274 less zero point 279 plus 0.2 to 1, which comes out to 0.775 now for Part B were given that there are two umbrellas on Sunday night at home. And what are the chances that there will be zero umbrellas to take to work on the following Thursday morning? So that is the same is asking the question. What is the probability that there will be zero umbrellas at home on Wednesday night after work. So that is three steps from Sunday night to Wednesday night. That means we're looking for the three step probability. I'm going from two umbrellas at home to know umbrellas at home. So this time, of course, we calculate we calculate P three, which is the three step transition matrix or the Markov chain. And then we look at Entry 31 which corresponds to going from 2 to 0. Just to make that clear would be going for entry three one, which would be this entry. Except that this is the single step transition matrix. So that would be the probability of going from two 20 umbrellas. But of course these air actually really three and call him one, and this comes out to 0.38 in last report. See, we're told that Sarah has two umbrellas at home at the start of the week and were asked what the expected number of umbrellas you'll have at home at the end of Monday and at the end of Tuesday is so when we look at the transition matrix, So road Rule three, which is corresponds to having two umbrellas. It is thebe probability distribution for how many umbrellas are going to be at home after the next transition. So if we have two umbrellas at home, the probability of having zero umbrellas at home in one step is zero. The probability of having one umbrella at home in one step is 0.14 and so on. So this role was their probability distribution for how many umbrellas we're going to have at home after one transition. And so the expected value of a random variable is the sum of X times probability of X, And the summation is overall possible values of X. And in this case, X goes from 0 to 4. That's the number of umbrellas. It is the state space for this Markov chain. So looking at that row of the transition matrix, the probability of getting zero is zero. The probability of one is 0.14 Probability of getting to is 0.62 probability, so that should be multiplied by two and then three, and probability of three is 0.24 and the probability of getting four is zero, and that comes out to zero. That comes out to 2.1, so we could call that sort of the expected number of umbrellas, um, after one transition, which happens to be Monday night. So that's the answer to the first question. And then the second question is, what is the expected number of umbrellas tohave on Tuesday night? If we start with two on Sunday night, so this time we proceed in a similar manner, except we need to use the two step transition matrix. So we calculate this, so I won't write it all out. I'll just give the pertinent row. 0.196 zero point 1736 zero point 4516 0.2976 and 0.576 So if we currently have two umbrellas at home in two steps, our probability distribution for the number of umbrellas at home looks like this role. And so that is our probability distribution for how many umbrellas we're going to have at home on Tuesday night. So what is the expected number of umbrellas at home on Tuesday night? So it zero times 0.196 It's one times 0.1736 two times 0.4516 and so on and this comes out to 2.2.

This question is all about determining whether a pair of events is independent. So let's look at part A rolling a pair of dice on observing, too, on the first ice and total on the dices equals no, and it was always looking ahead was do the things affect each other? So if we roll a two on the first ice, it is actually impossible to get a total of 10 cause even two plus sex is only eight. So if you don't if you roll something Highlander, too, you're probably more like unless it's a lot more likely to get a total of 10. So these are not independent things, because if you roller to first ice, then effects the chances of getting a total of 10 in this case, actually. So, he says, drawing one card from a regular deck of playing cards and having a red card on having an ace. So chance of being read. Obviously, half the deck chances of a face is wheeler forced punk or 52. So hey, the red cost. Those are for the chance of it being an ex. So if we get a red card, we've got a two and 52 chants. So it fresh we got to in 52 chance of a nice If we get a black art. We also have a to 50 chance Bourbonnais. So either way, they say we have the same chance of getting an age, whether it's red or black. So it doesn't. The fact that is red doesn't affect our chance of getting an X. So it's independent movement. Ponsi. It says that it's raining today. Andi Passing to use exam No, it's just a case of does the fact that it's raining affect your chance of passing exams, and I would definitely see no so independent. So probably says Rains Day and playing golf today. Now, if it's raining, does that affect your chances of playing golf now? I would say just using common sense if it's raining, that means a lower chance of playing golf. But people people are likely to change their decisions on Gulf dependent on the weather, so they're dependent events. If it's not raining, people more likely to play golf the last time, says completing today's homework. So mint on being on time for class. No, this one, you could potentially Are you either way because you could say that if you completed states homework, you're more likely to turn up the car sometime. If you haven't, you're more likely toe try and cram it and stay late. Or you could say that the homework assignment and your ability to get 1000 time punk dependent. But I would probably said these air dependent events and that's with some assumptions. So when you're putting that, definitely all your case and it will help against it. So just to recap, uh, for a we had not independent be independent. See independent D not independent on E.


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